$f(x, y) = \cos(x) + \sin(y)$ $\dfrac{\partial^3 f}{\partial x^3} = $
Taking a third order partial derivative is like taking a regular third order derivative. We take the partial derivative once, then we take another partial derivative, and then we take one more. $\dfrac{\partial^3 f}{\partial x^3} = \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ \dfrac{\partial f}{\partial x} \right] \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^3 f}{\partial x^3} &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ \cos(x) + \sin(y) \right] \right] \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ \dfrac{\partial}{\partial x} \left[ -\sin(x) + 0 \right] \right] \\ \\ &= \dfrac{\partial}{\partial x} \left[ -\cos(x) \right] \\ \\ &= \sin(x) \end{aligned}$ Therefore, $\dfrac{\partial^3 f}{\partial x^3} = \sin(x)$.